A solenoid 15 cm long has 300 turns of wire. A current of 5 A flows through it. What is the magnitude of the magnetic field inside the solenoid?
Correct answer: C. 4π x 10-3
- A. 75 x 10^7 T
- B. 60 x 10+3 T
- C. 4π x 10-3
- D. 750π x 10+3
Explanation
The magnetic field inside a solenoid is given by the formula:B=μ0nIwhere:B = Magnetic field (T)μ0 = Permeability of free space (4π×10−7n = Number of turns per unit length (N/L)I = Current (A)Given data:N=300N = 300N=300 turnsL=15L = 15L=15 cm = 0.15 mI=5I = 5I=5 AFirst, calculate n:n=NL=3000.15=2000 turns/mNow, calculate BBB:B=(4π×10−7)×(2000)×(5)Approximating π≈3.1416B≈4×3.1416×10−3
This question appeared on the UHS MDCAT 2014 paper, which you can sit online with every answer explained.
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About Magnetic Flux Density
Magnetic flux density measures the magnetic force acting on a current carrying conductor or moving charge in a magnetic field. Work includes the relations B equals F divided by IL and F equals qvB, the tesla as its SI unit, field direction and the distinction between flux density and magnetic flux.
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